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If the limit $y = \lim_{n \to \infty} f(x; n)$ exists, wouldn't it be given by $$y = x + \frac1{y} \implies y^2-x y-1=0 \implies y = \frac{x}{2} + \frac12 \sqrt{x^2+4}$$ Then, by dominated convergence, $$\lim_{n \to \infty} \int_0^1 dx \, f(x;n) = \frac12 \int_0^1 dx \, \left (x + \sqrt{x^2+4} \right ) = \frac{\phi}{2} + \log{\phi}$$ This is close to, ...
Interpret the map $$T:\ u\mapsto {1\over 2+u}\tag{1}$$ as a Moebius transformation of the complex Riemann sphere $\bar{\mathbb C}:={\mathbb C}\cup\{\infty\}$. Solving the equation $T(u)=u$ gives the two fixed points $$-1+\sqrt{2},\qquad -1-\sqrt{2}\ .$$ We therefore introduce a new complex coordinate $z$ on $\bar{\mathbb C}$ via $$z:={u+1-\sqrt{2}\over ... 2 Is easy to check that for x>\sqrt2-1, x>1/(2+x)>\sqrt2-1, so$$U_{n-1} >\sqrt2-1\implies \sqrt2-1<U_n = \frac1{2+U_{n-1}} < U_{n-1}.$$Starting from U_1>\sqrt2-1, this proves that the sequence is decreasing and bounded, so convergent. Now, take limits in U_n = \frac1{2+U_{n-1}}. 1 To properly solve this problem, first of all, why should the limit exist at all? As per any problem solving, we start with looking at a few numbers: First of all, the sequence itself is not well defined - we do not know where does the sequence start, or  U_1 , so let's just put some guess  U_1 = -100  to start with, the sequence of numbers look like ... 4 The method is to make a substitution U(n)=\frac{T(n)}{T(n+1)} and you would get much more tangible$$T(n+1)^2-2T(n)T(n+1)-T(n)^2=0$$This one you solve assuming T(n)=a^n and when you substitute and solve you have that a_1=1-\sqrt{2} and a_2=1+\sqrt{2} which gives$$T(n)=c(1-\sqrt{2})^n+d(1+\sqrt{2})^n and the solution follows. Set initial ...